I know that to find stationary points on a function, we need to differentiate the function and set that = 0.
But how can we find the stationary points of the below function?
$$y=\frac{1}{x} + \frac{1}{x^2} – \frac{1}{x^3}$$
When I differentiate it, I get fractions with $x$ in the denominator… I try to factorize and I get a quadratic equation with non-real roots and then $\frac{1}{x^4}$ which $= 0$. But the only way for $\frac{1}{x} = 0$ is if $x=\infty$.
I thought maybe there are no stationary points but this function does have stationary points that can be seen here:
https://www.wolframalpha.com/input/?i=1%2Fx+%2B+1%2Fx%5E2+-+1%2Fx%5E3
The maximum is at $x=1$ and the minimum at $x=-3$.
How can we solve this problem? Or more specifically, perhaps the question is how to solve the equation $\frac{1}{x^n} = 0$ ?
Best Answer
Hmm. \begin{align*} y&=\frac{1}{x}+\frac{1}{x^2}-\frac{1}{x^3} \\ &=\frac{x^2+x-1}{x^3} \\ y'&=\frac{x^3(2x+1)-3x^2(x^2+x-1)}{x^6} \\ &=\frac{x(2x+1)-3(x^2+x-1)}{x^4} \\ &=\frac{2x^2+x-3x^2-3x+3}{x^4} \\ &=\frac{-x^2-2x+3}{x^4}. \end{align*} Setting this equal to zero is tantamount to solving $x^2+2x-3=0,$ with solutions $$x=\frac{-2\pm\sqrt{4+4(3)}}{2}=\frac{-2\pm 4}{2}=\{1, -3\}. $$ You can probably use the Second Derivative Test to show which of these is a local min, and which a local max.